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Evaluate$$\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$$

Difficult problem. Been thinking about it for a few hours now. Pretty sure it's beyond my ability. Very frustrating to show that the limit even exists.

Help, please. Either I'm not smart enough to solve this, or I haven't learned enough to solve this. And I want to know which!

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just playing around, $\lim_{x \to 1^-}\left( \frac{\lim_{n \to \infty }\left( 1-x^{2n}\right) - (1-x^{2})\log_2 \left( 1 \over 1-x\right)}{1-x^{2}}\right)$, this seem to go like $\lim_{n \to \infty} n - \log_2 n$ –  Santosh Linkha Feb 6 '13 at 6:29

2 Answers 2

Write $x:=e^{-2^{\delta}}$. Then the desired limit is $\lim_{\delta\to-\infty} F(\delta)+\log_2 (1-e^{-2^{\delta}})$, where $$ F(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}.$$ But if $$ G(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}+\sum_{n<0} (e^{-2^{\delta+n}}-1) $$ then shifting the index of summation shows that $G(\delta+1)=G(\delta)-1$, so $G(\delta)+\delta$ has period $1$. Calling this periodic function $H(\delta)$, then, \begin{eqnarray*} F(\delta)+\log_2 (1-e^{-2^{\delta}}) &=& H(\delta) -\delta + \log_2 (1-e^{-2^{\delta}}) - \sum_{n<0} (e^{-2^{\delta+n}}-1)\\ &=&H(\delta)+O(2^\delta),\qquad\delta\to-\infty. \end{eqnarray*} Computing the periodic function $H$ numerically shows that it is not a constant. Therefore, the function whose limit is being taken is oscillatory, so the limit does not exist.

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This is NOT a solution, but I think that others can benefit from my failed attempt. Recall that $\log_2 a=\frac{\log a}{\log 2}$, and that $\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$ for $-1\leq x<1$, so your limit becomes

$$\lim_{x\to1^-}x+\sum_{n=1}^\infty\biggl[x^{2^n}-\frac1{\log2}\frac{x^n}n\biggr]\,.$$

The series above can be rewritten as $\frac1{\log2}\sum_{k=1}^\infty a_kx^k$, where

$$a_k=\begin{cases} -\frac1k,\ &\text{if}\ k\ \text{is not a power of}\ 2;\\\log2-\frac1k,\ &\text{if}\ k=2^m.\end{cases}$$

We can try to use Abel's theorem, so we consider $\sum_{k=1}^\infty a_k$. Luckily, if this series converges, say to $L$, then the desired limit is equal to $1+\frac L{\log2}\,$. Given $r\geq1$, then we have $2^m\leq r<2^{m+1}$, with $m\geq1$. Then the $r$-th partial sum of this series is equal to

$$\sum_{k=1}^ra_k=\biggl(\sum_{k=1}^r-\frac1k\biggr)+m\log2=m\log2-H_r\,,$$

where $H_r$ stands for the $r$-th harmonic number. It is well-known that

$$\lim_{r\to\infty}H_r-\log r=\gamma\quad\text{(Euler-Mascheroni constant)}\,,$$

so $$\sum_{k=1}^ra_k=\log(2^m)-\log r-(H_r-\log r)=\log\Bigl(\frac{2^m}r\Bigr)-(H_r-\log r\bigr)\,.$$

Now the bad news: the second term clearly tends to $-\gamma$ when $r\to\infty$, but unfortunately the first term oscillates between $\log 1=0$ (when $r=2^m$) and $\bigl(\log\frac12\bigr)^+$ (when $r=2^{m+1}-1$), so $\sum_{k=1}^\infty a_k$ diverges.

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Perhaps $\sum a_k$ is Cesàro summable? Abel's theorem would still apply then. –  Antonio Vargas Feb 6 '13 at 7:18
    
@AntonioVargas Thanks for the suggestion, though from David Moews' answer we can conclude that $\sum a_k$ is not Cesàro summable, either. By the way, where can I find a proof of this more general version of Abel's theorem? Almighty Wikipedia does not make any reference to this generalization. –  Matemáticos Chibchas Feb 10 '13 at 1:04
    
I learned the result from Hardy's book Divergent Series. Unfortunately I don't have a copy handy so I can't point you to a particular page. –  Antonio Vargas Feb 10 '13 at 2:02

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